# Finite Element Analysis Of A Cracked Plate Biology Essay

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The aim of this coursework is to learn the Linear Elastic Fracture Mechanics Analysis of a cracked plate using the Abaqus Finite Element Analysis software by predicting the stress intensity factors of the edge-cracked plate in under various conditions i.e. variations of cracked length, variations of cracked dimensions/locations, variations of loading condition, variations of materials, variations of plate dimensions, variations of element type, numbers of contours etc.

In industries, it is observed that the failures of engineering components/structures are because of fractures by cracks propagations. The cracks propagation depends on the type of loading condition. There are two kind of cracks i.e. one is stable & another is unstable. The intension of this coursework is to study the stress intensity factor of edged cracked plate individually & in Group using the Abaqus FEA software. In individual work, the parameters of plate dimensions, crack dimensions/locations & loading conditions are changed in order to see the effects of the stress intensity factors using Abaqus software. Similarly in group work, the analysis in Abaqus is performed considering the different element types i.e. quadratic triangular & quadrilateral and different contours in order to see the effects of the stress intensity factors of the edge-cracked plate.

## Theory:

Fracture mechanics is the subject of study for cracks & crack type of defects in engineering components/structures. The aim of the Fracture mechanics is to study the behaviour of the cracks which lead to fracture & also the tolerance of the components/structures. It is also concerned with the cracks propagates & grow under various loading conditions. There are two types of cracks.

Stable cracks (which are slow & safe).

Unstable cracks (which are instantaneous and catastrophic).

In many fractures problems, the fracture analysis is made considering pure elastic behaviour of the material using linear fracture mechanics (LEFM). However, there are the situations where significant plastic deformation takes place in the vicinity of the tip, and yielding fracture mechanics is to be considered.

There are three kinds of Fracture modes.

Opening (tensile) mode: The crack surfaces move directly apart. This is referred to as mode I.

Sliding (Shearing) mode: The crack surfaces slide over one another in the direction perpendicular to the leading edge of the cracks. This is referred to as mode II.

Tearing mode: The crack surfaces move relative to one another and parallel to the leading edge of the cracks. This is referred to as mode III.

The stress intensity factor (SIF) is non-dimenstionalised as below for mode I.

## KI = KI / Ïƒ(Ï€a)^1/2

Where,

KI = The non-dimenstionalised stress intensity factor.

KI = The stress intensity factor in MPa.m1/2.

a = The crack length in metre.

The singular elastic stress distributions in the immediate vicinity of a crack tip can be completely characterised by a stress intensity factor. The stress intensity factor depends on the loading & geometry of the body & crack.

The Finite element method is applied to 2 dimensional stress problems using two methodologies of meshing.

Linear interpolation

Linear interpolation: In linear interpolation meshing, the triangle element type having 3 nodes is used for analysis. This method is more appropriate involving linear variations of the displacements & stresses over the solution domain. The each side of the element remains straight. This kind of element meshing can't be applied to the problem when dealing with the curved shape sides.

Quadratic interpolation: In quadratic interpolation meshing, there are two kinds of elements are used.

Triangular element: In this kind of element, the 6 nodes are used i.e. 3 nodes at vertex & 3 nodes at mid of the each side.

Quadrilateral element: In this kind of element, the 8 nodes are used i.e. 4 nodes at the each corner & 4 at mid of the each sides.

The sides of quadratic type of elements are uniquely defined by quadratic functions of position passing through the 3 nodes along each sides which automatically satisfy the compatibility both within each element & the inter element boundaries. This type of meshing element is also known as higher order elements. The advantage of higher order elements is that the shape function is capable of representing the true variation more accurately. This type of meshing is more appropriate when dealing with curved shape which gives accurate results. It also provides good compromise between flexibility of element shape and accuracy of shape function & also the amount of the computation required.

## Finite element analysis treatment of Fracture problems:

It is very essential to use relatively large numbers of relatively small elements (quadrilateral quadratic elements) in the region of the stress concentration in order to resolve the varying stress field. Typical crack analysis use quadrilateral elements for 2 dimensional problems & brick elements for 3 dimensional problems.

## The mid side node of the quadratic elements is always shifted by 1/4 to the quarter points nearest at the crack tip in order to keep the stress singular at the crack tip for accurate numerical results.

Individual work: In this course work, the six contours have been considered for accurate results of stress intensity factor at the crack tip. The accuracy of the stress results depends on the number of contours i.e. if numbers of contours increase, the accuracy also increases. The triangular quadratic elements are used in meshing for the small contour (inner most circles) because quadrilateral elements always collapsed down into triangles occupying the same points in the space which is more appropriate to fit the curved shape & outside of this contour along with the whole plate; quadrilateral quadratic elements are used for accurate result.

Group work: In this course work, the different contours are used i.e. from 1 to 6 having the same radius as used in individual work in order to see the effects of the stress at the crack tip. In this course work the smallest counter is considered the tri element. The element area selected for the rest contours except small contour is considered the quad by changing the plate element tri & quad alternately. This gives the different results of the stress intensity factors at the crack tip.

## Procedure:

For Group & Individual:

Module: Part: Create Part

Enter the Name for the part (optional)

Select 2D Planner, Deformable & Shell

Enter 20 for Approximate size.

Create lines: Rectangle. The rectangle can be drawn by entering x &y coordinates for the starting corner & end corner as per the different case numbers depending on the size of the plate.

## Module: Property:

Create property: Enter the Name of the material (optional). Click the Mechanical by choosing Elasticity & elastic. Enter the Young's modulus (E) & Poisson's Ratio depending on the types of materials considered in the course work as mentioned below.

Young Modulus:

Steel=200e9(GPa)

Aluminium= 70e9(GPa)

Copper=120e9(GPa)

Poisson's Ratio:

Steel=0.3

Aluminium= 0.33

Copper=0.36

Create section: Enter Name (optional), choose solid, Homogeneous. Enter the thickness 1 for Group work & 0.1 for Individual work.

Assign section: Select the part by clicking it & click the done to assign the section.

## 3) Module: Assembly:

- Click the instance part choosing Independent (mesh on instance) & click ok.

## 4) Module: Step:

- Enter a Name (optional), Select static, General & click on continue. Click ok on new open window.

5) Module: Assemble: (To create crack line)

- Click on Partition Face: sketch, click on Created Lines: connected. Enter the co-ordinates of the starting point & end point depending on the crack length & location as per different cases of group & individual work. Press escape to end the command & click on done which will create the crack line.

## 6) Module: Interaction:

- Click on Special from menu bar & choose the crack & click on assign Seam by clicking on the drawn crack line & click done.

- Choose on Special, choose Crack & click on Create. Enter a Name (optional). Click continues. Select the front by clicking on the crack tip point. Click on done. Click on q vectors: Enter the co-ordinates for start & end point to create the direction of crack propagation depending on the crack length & location as per the cases of course work.

The new window will open. Click on Singularity tab. Enter the 0.25 for midside node parameter. Select collapsed element side, single node. Click ok.

7) Module: Step:

- Click on History Output Manager. Click on Edit. New window will open. Click on the drop menu of Domain: and choose integral. Enter 6 for Number of contours. Select Stress Intensity Factors & click ok. Click on Dismiss in the History Output Requests Manager window.

- Enter a Name (optional). Click on Pressure. Also concentrated force for individual course works of case no 11 & 12. The magnitude of the pressure/point load is considered as given in the course work. Click on Continue. Select the lines/points for the loads to be applied depending on the types of loading. Enter the respective values of the load. Click on ok.

## 9) Module: Mesh:

- Click on Partition Face: sketch. Click on sketch option icon to align the geometry of the plate with the grid.

Remove the tick from auto in front of 'Grid spacing'. Type1 for group work &1.29 for individual work in the 'Grid spacing' box. Click ok. Click on create lines Rectangle. Click on bottom left corner and draw a 1 unit square for group work & 1.29 for individual work by using 'Add dimension' icon to confirm the correct dimension. Click on the 'Linear pattern' icon & select the right edge & top edge to create pattern. Click done. New window will open. Type in boxes Number & spacing for direction 1 & 2. This depends on the geometry of plate & crack. For group work, type Number-10, spacing 1 for Direction 1 and Number 20, spacing 1 for Direction 2. Click ok. Click done to end partitioning.

-Select the partition. Select the singular region around the crack tip by selecting the four squares around the crack tip. Click on done. Click on the construction icon & choose the horizontal line passing through the crack tip. Draw diagonal lines through crack tip. Draw the 6 circles centred at the crack tip with radii: 0.6, 0.36, 0.216, 0.13, 0.078 and 0.047 which depends on the number of contours considered as given in the course work. The number of circles depends on the number of contours considered. For example, if one contour then select the one circle having the small radius. As the number of contours increases, the number & radius of the circles will also increase. The circle can be drawn using Create Circle icon & also Add dimension icon for entering the value of radii. Press enter. Press escape to end any command in use. Now plate is portioned.

- Click on Seed Part Instance. Enter 1 for group work & 1.29 for individual work for approximate global size. Click on apply & then ok.

## - Click on Assign Mesh Controls.

For Group work:

Select the whole plate & select the element type tri or quad as given in the Group course work. Click ok. Now select the element around crack tip always tri i.e. innermost circle. Select the quad element for square block for singular region around crack tip except innermost circle. Click ok.

For Individual work:

Select the whole plate & select the element type quad as given in the Group course work. Click ok. Now select the element around crack tip always tri i.e. innermost circle. Select the quad element for square block for singular region around crack tip except innermost circle. Click ok.

## Click on Assign Element type:

Select the whole plate. Click on Done. Select the following options: standard (in Element library), Quadratic ( in Geometric order), plane Stress (in Family), and unselect Reduced integration. The element name should be displayed automatically as CPS8. Click on the tri tab and unselect modified formulation. Click ok.

Click on Mesh Part instance & click yes. The plate is meshed.

## 10) Module: Choose Job:

- Click on job manager. Click on create. Enter the name for the job (optional). Click continue. Click ok in new window. Click on submit. Once it displays completed analysis under status, click on the results to see the deformed shape. Click on the deformed shape to view the deformed shape of the plate.

## 11) Module: Visualisation:

- To view the stress intensity factors (SIFs), click on Results in the menu bar & click on History Output. A new window will open. Select the SIFs for mode I & II & plot them for each contour.

- Note down the values of SIFs for each contour & prepare the table for different case as given in the course work for further study /analysis of the results.

## For Group work:

Table-1 (Input Table for Group work)

Case No.

w(m)

H(m)

a/w

a(m)

h/H

h(m)

Ïƒ(MPa)

Material

Element type

Number of contours

1

10

20

0.5

5

0.5

10

1.0

steel

Tri

1

2

10

20

0.5

5

0.5

10

1.0

steel

1

3

10

20

0.5

5

0.5

10

1.0

steel

Tri

2

4

10

20

0.5

5

0.5

10

1.0

steel

2

5

10

20

0.5

5

0.5

10

1.0

steel

Tri

3

6

10

20

0.5

5

0.5

10

1.0

steel

3

7

10

20

0.5

5

0.5

10

1.0

steel

Tri

4

8

10

20

0.5

5

0.5

10

1.0

steel

4

9

10

20

0.5

5

0.5

10

1.0

steel

Tri

5

10

10

20

0.5

5

0.5

10

1.0

steel

5

11

10

20

0.5

5

0.5

10

1.0

steel

Tri

6

12

10

20

0.5

5

0.5

10

1.0

steel

6

## Notations:

w=Width of the plate in metre

H=Height of the plate in metre

a=Length of the crack tip in metre

h=Distance of the crack tip from bottom of the plate in metre

Ïƒ=Applied stress/pressure on the plate in MPa.

K1 =Stress intensity factor for mode I i.e. opening mode in MPa.m1/2

K2 =Stress intensity factor for mode II i.e. shearing mode in MPa.m1/2

K1=Non dimenstionalised Stress intensity factor for mode I i.e. opening mode.

K2=Non dimenstionalised Stress intensity factor for mode II i.e. shearing mode.

F1=Applied load on the plate in KN.

F2=Applied load on the plate in KN.

Table-2 (Result Table for Group work)

14.14

0.00647

3.57

14.14

-6.87E-04

3.57

12.60

12.37

4.98E-03

5.03E-03

3.12

12.61

12.38

3.34E-05

0

3.12

11.52

11.64

11.64

4.93E-03

4.87E-03

4.92E-03

2.94

11.86

11.64

11.64

0

0

0

2.94

11.52

11.30

11.30

11.30

4.93E-03

4.86E-03

4.86E-03

4.90E-03

2.91

2.85

2.85

2.85

11.52

11.31

11.31

11.31

5.12E-05

-1.34E-05

-1.73E-05

-2.23E-05

2.85

11.40

11.12

11.12

11.12

11.12

4.91E-03

4.82E-03

4.82E-03

4.82E-03

4.85E-03

2.82

11.40

11.12

11.12

11.12

11.12

2.48E-05

0

0

0

-1.40E-05

2.82

11.37

11.16

11.16

11.16

11.16

11.16

4.77E-03

4.68E-03

4.68E-03

4.68E-03

4.68E-03

4.70E-03

2.82

11.38

11.17

11.17

11.17

11.17

11.17

1.43E-05

0

0

0

0

0

2.82

## For Individual work:

Table-3 (Input Table for Individual work)

Case No.

w(m)

H(m)

a/w

a(m)

h/H

h(m)

Ïƒ(MPa)

F1(KN)

F2(KN)

Material

Number of contours

1

12.9

25.8

0.1

1.29

0.1

2.58

1.0

0

0

steel

6

2

12.9

25.8

0.1

1.29

0.5

12.9

1.0

0

0

steel

6

3

12.9

25.8

0.5

6.45

0.1

2.58

1.0

0

0

steel

6

4

12.9

25.8

0.5

6.45

0.5

12.9

1.0

0

0

steel

6

5

12.9

25.8

0.5

6.45

0.5

12.9

2.0

0

0

steel

6

6

12.9

25.8

0.5

6.45

0.5

12.9

5.0

0

0

steel

6

7

12.9

64.5

0.5

6.45

0.5

12.9

1.0

0

0

steel

6

8

12.9

129

0.5

6.45

0.5

12.9

1.0

0

0

steel

6

9

12.9

25.8

0.5

6.45

0.5

12.9

1.0

0

0

aluminium

6

10

12.9

25.8

0.5

6.45

0.5

12.9

1.0

0

0

copper

6

11

12.9

25.8

0.5

6.45

0.5

12.9

1.0

100

100

steel

6

12

12.9

25.8

0.5

6.45

0.5

12.9

1.0

250

250

steel

6

Table-4 (Result Table for Individual work)

2.97

2.92

2.92

2.92

2.92

2.92

0.165

0.162

0.162

0.162

0.162

0.162

1.45

0.081

2.44

2.40

2.40

2.40

2.40

2.40

1.19

20.33

19.95

19.95

19.95

19.95

19.95

6.46

6.35

6.35

6.35

6.35

6.35

4.43

1.41

12.93

12.69

12.69

12.69

12.69

12.69

0

0

0

0

0

0

2.82

25.86

25.39

25.39

25.39

25.39

25.39

0

0

0

0

0

0

2.82

64.65

64.47

64.47

64.47

64.47

64.47

0

0

0

0

0

0

2.82

11.93

11.69

11.69

11.69

11.69

11.69

-19.75E-06

0

0

0

0

14.33E-06

2.82

12.93

12.69

12.69

12.69

12.69

12.69

-25.93E-06

0

0

0

14.56E-06

18.80E-06

2.82

12.93

12.69

12.69

12.69

12.69

12.69

-3.26E-05

0

0

1.42E-05

1.84E-05

2.37E-05

2.82

12.93

12.69

12.69

12.69

12.69

12.69

0

0

0

0

0

0

2.82

16.12

15.82

15.82

15.82

15.82

15.82

0

0

0

0

0

0

2.82

20.90

20.52

20.52

20.52

20.52

20.52

0

0

0

0

0

0

4.56

## Note:

The width of the plate & height of the plate is multiplied by x dimension where x dimension is 129mm which is unique for each student for individual work. The thickness of the plate is 100mm.

The Tri element is used at crack tip & Quad elsewhere.

## For Group Work:

In the Group work, the edge-cracked plate analysis is carried out using Abaqus software by changing the element type i.e. tri & quad & the numbers of contours as mentioned in table 1 (input table for group work) inorder to see the effect on the stress intensity factors at cracked tip & the results are prepared in tabular form in table 2(result table for group work). For illustrating the effects of the variation in said FEM parameters, the case 1, 2, 5,6,11 & 12 are considered & explained the results as below with deformed figures/graphs as necessary.

Note: The course work for edge-cracked tip plate is subjected to only opening mode as the nature of applied load/stress is tensile.

## Discussion on Group work results:

By observing the result table 2, it is noted that the value of stress intensity factor K1 (opening mode I) is more near to crack tip & becomes constant as moving away from the crack tip. This can be easily understood by graphs for case-11 & 12 as shown above.

It is also observed from the result table that tri element type meshing gives values of stress intensity factors for opening mode (K1) & also for shear mode (K2). The value of K2 is very less i.e. negligible or equivalent to 0. As this is the symmetric kind of problem where the value of K2 must be 0 theoretically but it is not possible in tri element type meshing. The value of K2 is found 0 in quad type element which proves that the accuracy of quad type element is more rather than tri type element.

It is noted that the value of stress intensity factor K1 is more for case 1 & 2(i.e. 14.14 MPa.m1/2) which is having only 1 contour & the same is less for case 11 & 12(i.e. 11.38 MPa.m1/2) which are having 6 contours. This proves that the accuracy of the results for stress intensity factors depends on the number of contours. As the number of contours increases, the accuracy of the result also increases.

## For Individual Work:

In the Individual work, the edge-cracked plate analysis is carried out using Abaqus software by changing the dimension of the plate, dimension of the crack tip, location of the crack tip, load condition & materials as mentioned in table 3 (input table for individual work) inorder to see the effect on the stress intensity factors at cracked tip & the results are prepared in tabular form in table 4(result table for individual work). For illustrating the effects of the variation in said parameters, the case 1, 2, 3,6,7,8,9,10,11 & 12 are considered & explained the results as below with deformed figures/graphs as necessary.

Note: The course work for edge-cracked tip plate is subjected to only opening mode as the nature of applied load/stress is tensile.

## Discussion on Individual work results:

By observing the result table 4 for individual work, it is noted that the value of stress intensity factor K1 (opening mode I) is more near to crack tip & becomes constant as moving away from the crack tip. This can be easily understood by graph-3 for Case-12(individual work) as shown above.

By observing the deformed figure for case-1 & 2, it is seen that crack tip with the same length (1.29m) is not symmetry in case -1 but it is symmetry in case-2.The results obtained from the analysis prepared in the table-4, it is seen that the symmetry type crack tip gives the value of stress intensity factors for opening mode but it is not applicable for shearing mode because of the nature of geometry. Whereas the non symmetry type crack tip gives the value of stress intensity factors for both mode i.e. opening mode & shearing mode. This shows that the shearing takes place in non symmetry type crack tip but the value of K2(shearing mode) is small as compare to K1(opening mode). This proves that non symmetry type crack tip will also give some sort of shearing.

## Hence the stress intensity factors depend on the geometry of the crack tip i.e. symmetry or non symmetry.

Example: For case-1, the value of K1= 2.97 & 2.92 in MPa.m1/2 and K2= 0.165 & 0.162 in MPa.m1/2

For case-2, the value of K1=2.44 & 2.39 in MPa.m1/2 and K2= 0.

By observing the deformed figure for case-3 & result table-4, it is seen that the value of the SIFs(stress intensity factors) is more i.e. K1= 20.32 & 19.95 in MPa.m1/2 for 6.45m crack tip length which is quite more in the case of 1 & 2. This proves that as the length of the crack tip increases, the values of SIFs also increases so the stress intensity factors depend on the length of the crack tip.

By observing the deformed figure for case-6 & result table-4, it is seen that the value of SIFs is more rather than case-4 i.e. for case-6: K1= 64.65 & 64.47 in MPa.m1/2 for 6.45m crack tip length & 5MPa applied stress & for case-4: K1= 12.93 & 12.69 in MPa.m1/2 for 6.45m crack tip length & 1MPa applied stress. This difference occurs because of the applied stress. The applied stress is more in the case-6 rather than case-4 which resulted drastic change in the values of SIFs (K1).

## Hence the value of SIFs depends on the applied stress as the stress increases, the SIFs also increases.

By observing the deformed figure for case-7 &8 and result table-4, the values of SIFs (K1) for both the cases are same i.e. K1= 12.93 & 12.69 in MPa.m1/2 for different length i.e. for case-7: 64.5 m & for case-8: 129 m.

## Hence the value of SIFs (K1) does not depend on the length of the plate.

By observing the deformed figure for case-4,9 &10 and result table-4, the values of SIFs (K1) for cases 4,9 & 10 are same i.e. K1= 12.93 & 12.69 in MPa.m1/2 for different materials i.e. for case-4: steel material & for case-9: aluminium material & for case-10: copper material

## Hence the value of SIFs (K1) does not depend on the materials

By observing the deformed figure for case-11 &12 and result table-4, the values of SIFs (K1) are more for case-12 while comparing with case-11. This difference occurs because of the concentrated point load as mentioned in the input table-3.

For case-11: K1= 16.12 & 15.82 in MPa.m1/2 for 100 KN of F1 & F2

For case-12: K1= 20.90 & 20.51 in MPa.m1/2 for 250KN of F1 & F2

## Conclusions:

In the view of the above results, following conclusions are made.